Using prime factorization method, find which of the following numbers are perfect squares?

Question:

Using prime factorization method, find which of the following numbers are perfect squares?

189, 225, 2048, 343, 441, 2916, 11025, 3549

Solution:

(i) 189 = 3 x 3 x 3 x 7

Grouping them into pairs of equal factors:

189 = (3 x 3) x 3 x 7

The factors 3 and 7 cannot be paired. Hence, 189 is not a perfect square.

(ii) 225 = 3 x 3 x 5 x 5

Grouping them into pairs of equal factors:

225 = (3 x 3) x (5 x 5)

There are no left out of pairs. Hence, 225 is a perfect square.

(iii) 2048 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2

Grouping them into pairs of equal factors:

2048 = (2 x 2) x (2 x 2) x (2 x 2) x (2 x 2) x (2 x 2) x 2

The last factor, 2 cannot be paired. Hence, 2048 is not a perfect square.

(iv) 343 = 7 x 7 x 7

Grouping them into pairs of equal factors:

343 = (7 x 7) x 7

The last factor, 7 cannot be paired. Hence, 343 is not a perfect square.

(v) 441 = 3 x 3 x 7 x 7

Grouping them into pairs of equal factors:

441 = (3 x 3) x (7 x 7)

There are no left out of pairs. Hence, 441 is a perfect square.

(vi) 2916 = 2 x 2 x 3 x 3 x 3 x 3 x 3 x 3

Grouping them into pairs of equal factors:

2916 = (2 x 2) x (3 x 3) x (3 x 3) x (3 x 3)

There are no left out of pairs. Hence, 2916 is a perfect square.

(vii) 11025 = 3 x 3 x 5 x 5 x 7 x 7

Grouping them into pairs of equal factors:

11025 = (3 x 3) x (5 x 5) x (7 x 7)

There are no left out of pairs. Hence, 11025 is a perfect square.

(viii) 3549 = 3 x 7 x 13 x 13

Grouping them into pairs of equal factors:

3549 = (13 x 13) x 3 x 7

The last factors, 3 and 7 cannot be paired. Hence, 3549 is not a perfect square.

Hence, the perfect squares are 225, 441, 2916 and 11025.